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Alex787 [66]
3 years ago
7

Prepare 10.00 mL of 0.010 M NaOH by diluting the NaOH solution used in Trial 1. (Your procedure should clearly show your calcula

tions and the glassware used to perform the dilution.)
Chemistry
1 answer:
denis-greek [22]3 years ago
3 0

Answer:

<em> = 0.2 mL.</em>

Explanation:

Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:

C_{1} V_{1} = C_{2} V_{2}

where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.

Let C1 = 0.5M, V2 = ?

C2 = 0.010M; V2 = 10mL

⇒Volume of stock solution to be diluted, V2

= \frac{10}{0.5} × 0.010

<em> = 0.2 mL.</em>

Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.

I hope this was a detailed explanation given the missing details of "Trial 1" in the question.

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          Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

                  Volume = length × width × height

                                =  (36.0 × 24.0 × 5.0) cm^{3}

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or,                             = 4.32 \times 10^{3} cm^{3}

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

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Hence, mass of displaced water is as follows.

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Therefore, calculate the density of Styrofoam as follows.

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